# gameroom.io blog

3Mar/115

## The Dominion of Dukes

For a quick introduction to the Duke and the game Dominion, click show more.

Every time the Duke makes an appearance in Dominion, I try to justify putting him in my deck. But in the hubbub of a live game, I never find time to work out precisely how many Dukes and Duchies I’d need to beat someone going for Provinces, or what distribution would best serve me.

Let’s use a little math and figure it out.

We’re going to determine:

1. The ideal balance of Dukes and Duchies given a fixed number of $5-purchases. 2. How many victory points that balance yields, so we can compare this strategy against others. ## The Duke-Duchy Balance We could brute force it easily, but where’s the fun in that? Instead, we’re going to do some simple calculus. Each Duchy is worth $$3$$ victory points, and each Duke is worth $$1$$ victory point per Duchy. We’re trying to maximize victory points $$V$$, under the constraint of a fixed number of$5-purchases $$p$$.

For $$k =$$ the number of Dukes and $$c =$$ the number of Duchies, $$V = 3c + kc$$ and the constraint is $$p = k + c.$$ We’ve fixed $$p$$ because we assume we have a rough idea how many 5-purchase opportunities we’re going to get before the game ends — we’re just deciding what ratio of Dukes to Duchies we’ll buy with those opportunities. Let’s now maximize $$V$$! \begin{align} k & = p – c \\ V & = 3c + (p – c)c \\ V & = 3c + pc – c^2 \end{align} For fixed $$p$$, $$V$$ is maximized at \begin{align} \frac{\partial V}{\partial c} & = 3 + p – 2c = 0 \\ c & = \frac{3 + p}{2}. \end{align} That’s fine for odd $$p$$, but for even $$p$$, the maximal $$c$$ is a non-integer, so it isn’t a possible quantity of Duchy cards. We’ll want one of the two nearest integer points, $$c_\pm = \frac{3 + p}{2} \pm \frac{1}{2}$$ but which one? Fortunately, $$V$$ is a parabola, symmetric around its maximum $$c = \frac{3 + p}{2}$$. Since both points are off by the same amount, $$\frac{1}{2}$$, both points have the same value — they both maximize $$V$$. So, for $$p$$ odd, \begin{align} c & = \frac{p+3}{2} \\ \end{align} and for $$p$$ even, \begin{align} c_\pm & = \frac{p+3}{2} \pm \frac{1}{2} \text{ (both work)}. \end{align} Let’s finish up our answer by putting it in terms of Dukes and Duchies. For $$p$$ odd, \begin{align} c & = \frac{c+k+3}{2} \\ 2c & = c + k + 3 \\ \\ c & = k + 3 \\ \end{align} And for $$p$$ even, \begin{align} c_\pm & = \frac{c_\pm + k_\pm + 3}{2} \pm \frac{1}{2} \\ 2c_\pm & = c_\pm + k_\pm + 3 \pm 1 \\ \\ c_\pm & = k_\pm + 3 \pm 1 \\ \end{align} So, for $$p$$ odd we want $$3$$ more Duchies than Dukes, and for $$p$$ even we want as close to that as possible. ### A discrete framing We can motivate this result with a discrete framing of the problem: When you buy a Duke, you get $$c$$ victory points. When you buy a Duchy, you get $$k + 3$$ victory points. So, when $$c > k + 3$$, you’re better off buying a Duke. When $$c < k + 3$$, you're better off buying a Duchy. And, naturally, when $$c = k + 3$$, you can buy either one. This implies that, once you have your first $$3$$ Duchies, you’re best off with $$c = k + 3$$ or as close to that as you can get. ## Maximal victory points How many victory points does the ideal Duke-Duchy balance yield? For $$p$$ odd, \begin{align} c & = k + 3 \\ \\ V & = 3c + kc \\ V & = c(k+3) \\ V & = c^2 \\ \end{align} and for $$p$$ even, \begin{align} c_\pm & = k_\pm + 3 \pm 1 \\ \\ c_+ & = k_+ + 4 \\ V & = c_+(3+k_+)\\ V & = c_+(c_+ -1 ).\\ \\ c_- & = k_- + 2\\ V & = c_-(3+k_-)\\ V & = c_-(c_- + 1).\\ \end{align} Note that because $$c_+ = c_- + 1$$, we see explicitly that $$V$$ is the same for $$c_+$$ and $$c_-$$. \begin{align} V = c_\pm (c_\pm \mp 1) \end{align} That’s nice, but it’s strategically useful to have this in terms of $$p$$, as a good player can tell roughly how many5-purchases she’s going to get before the game ends.

For odd $$p$$, \begin{align} c & = k + 3 \\ k & = p – c \\ c &= p – c + 3 \\ 2c &= (p+3) \\ c &= (p+3)/2 \\ V &= \frac{(p+3)^2}{4} \end{align} and for even $$p$$, \begin{align} c_\pm & = k_\pm + 3 \pm 1 \\ k_\pm & = p – c_\pm \\ c_\pm & = p – c_\pm + 3 \pm 1 \\ 2c_\pm & = p + 3 \pm 1 \\ c_\pm & = \frac{p + 3 \pm 1}{2} \\ \\ V & = \frac{p + 3 \pm 1}{2} \left( \frac{p+3 \pm 1}{2} \mp 1 \right) \\ V & = \frac{(p + 3 \pm 1) (p+3 \mp 1)}{4} \\ V & = \frac{(p + 4)(p+2)}{4}. \\ \end{align}

Let’s use these formulas to make a graph for quick reference during a game:

Show data table.

In a game where everyone’s doing about the same, the first player to $$4$$ Provinces usually wins. That’s worth $$24$$ victory points, so we see that you need $$7$$ $5-purchases to beat that player (barely). Even if your opponents can get more than $$4$$ Provinces (after all, there are more available given that you’re not buying any), once you reach $$p=7$$ you’ll keep pace with your Province-purchasing opponents if you simply match them buy-for-buy, pulling slightly ahead of them again at $$p=11$$ when they buy their $$8^{\text{th}}$$ Province. It’s noteworthy that because you need more Duchies than Dukes, you should bias yourself in the early game toward Duchies over Dukes beyond the $$c=k+3$$ that maximizes victory points. There are two reasons for this: 1. If an opponent decides to do the Duke-Duchy strategy, Duchies are going to run out before Dukes. 2. Province-purchasers will snap up the occasional Duchy on their own, both for the VPs and to keep it out of your Duke deck. This is especially true in 2-player, where denying an opponent the ability to score is just as valuable as scoring yourself. ## TL;DR In summary, the ideal Duke-Duchy strategy ends the game with $$3$$ more Duchies than Dukes. With $$p$$$5-purchases, this yields: \begin{align} V & = \frac{(p + 4)(p+2)}{4} \text{ for }p\text{ even,} \\ V &= \frac{(p+3)^2}{4} \text{ for }p\text{ odd.} \end{align}

So, to beat a player who has $$4$$ Provinces, the ideal strategy needs $$p=7$$ $5-purchases ($$5$$ Duchies and $$2$$ Dukes). To beat a player who has $$8$$ Provinces, the strategy needs $$p=11$$$5-purchases ($$7$$ Duchies and $$4$$ Dukes).